3
$\begingroup$

If $f:\mathbb{R} \to \mathbb{R}$ is a continuous function such that $f(x)=x$ has no real solution, then show that $f(f(x))=x$ has no real solution either.

Is the proof trivial as it seems or does it need an analytical approach?

$\endgroup$
2
  • 1
    $\begingroup$ Do you want a hint or a full solution? My hint is look at $f(x)-x$. $\endgroup$ – Emre Jun 4 '16 at 11:16
  • $\begingroup$ @Emre I need more than that, thanks. $\endgroup$ – StubbornAtom Jun 4 '16 at 11:17
9
$\begingroup$

As $f$ is continuous, $g(x)=f(x)-x$ is also continuous. We know that $g(x)=0$ has no root. Thus, either $g(x)>0$ for all $x\in\mathbb{R}$ or $g(x)<0$ for all $x\in\mathbb{R}$. In the first case, we have $f(x)>x$ for all $x\in\mathbb{R}$. So, $$f(f(x))>f(x)>x$$ In the second case, we have $f(x)<x$ for all $x\in\mathbb{R}$. So, $$f(f(x))<f(x)<x$$

$\endgroup$
2
  • $\begingroup$ I am having confusion. In your argument are you not assuming that the function is monotonically increasing when you say $f(x)>x \implies f(f(x))>f(x)$? $\endgroup$ – tomriddle99 Sep 13 '19 at 15:59
  • 1
    $\begingroup$ No, I am replacing $x$ with $f(x)$ in the statement. $\endgroup$ – Emre Mar 30 '20 at 2:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.